MIT PRIMES/Art of Problem Solving
CROWDMATH 2024: Where Does the Goldbach Conjecture Hold?
Resources
The Resources for this project are background ideas that you will need to understand and make progress on the CrowdMath problems. We've chosen resources that are directly relevant to the project: the problems are defined explicitly in terms of ideas that you will find in our resources. All of the problems we will pose are thematically linked and all of the resources we post will be as well. We will be releasing the resources roughly in the order that you should be reading them.
Each resource also has exercises to help clarify the key ideas and give practice with them. You can discuss the exercises by clicking either the "View Discussions" or "Start New Topic" button. You can discuss any resource by clicking the V or t buttons on the resources.
The Goldbach Conjecture
The Goldbach conjecture, introduced by Goldbach in a 1742 letter to Euler, remains one of the most ancient and renowned unsolved problems in mathematics. Modernly stated, the conjecture asserts that every even integer greater than can be expressed as the sum of two prime numbers.
Analogues of the Statements of Goldbach Conjecture
Since the 1960s, several authors have explored analogues of the Goldbach conjecture for various classes of polynomial rings. In 1965, Hayes demonstrated that polynomials with can be represented as the sum of two irreducibles. This work was later extended by Saidak, Kozek, and Pollack. Additionally, variations of the Goldbach conjecture for polynomials over finite fields have been investigated by Effinger and Hayes, Bender, Car and Gallardo, among others. More recently, Paran examined a version of the Goldbach conjecture in the ring of formal power series over the integers.
Semirings and Semidomains
A commutative semiring is a nonempty set endowed with two binary operations denoted by `' and `' and called addition and multiplication, respectively, such that the following conditions hold:
(1) is a monoid with its identity element denoted by ;
(2) is a commutative semigroup with an identity element denoted by with ;
(3) for all .
Let be a commutative semiring. Since the operation of addition in is cancellative, the distributive law ensures that for all . Throughout our CrowdMath project, for any , we write instead of when there seems to be no risk of ambiguity. In the typical definition of a `semiring' , one does not assume that the semigroup is cancellative. However, we do so here because our semirings of interest have cancellative additive structures. On the other hand, it is worth observing that a more general notion of a `semiring' does not assume that the semigroup is commutative. However, once again, this more general type of algebraic objects are not of interest in the scope of this paper. Accordingly, from now on we will use the single term semiring to refer to a commutative semiring, tacitly assuming the commutativity of both operations. A subset of is a subsemiring provided that is a submonoid of that is closed under multiplication and contains . Observe that every subsemiring of is a semiring. As an exercise, you will show that the set (resp., ) consisting of all polynomials (resp., Laurent polynomials) with coefficients in is also a semiring under the standard addition and multiplication of polynomials (resp., Laurent polynomials). We call (resp., ) the semiring of polynomials (resp., semiring of Laurent polynomials) over .
Definition. We say that a semiring is a semidomain provided that is a subsemiring of an integral domain.
Let be a semidomain. Then is a monoid, which we denote by and call the multiplicative monoid of . In the next example, we show a semiring that is not a semidomain but still its subset of nonzero elements is a multiplicative monoid. In order to reuse notation from ring theory, we refer to the units of the monoid as invertible elements of the semidomain , so that we can refer to the units of the multiplicative monoid simply as units of , avoiding any risk of ambiguity. Also, following standard notation from ring theory, we let denote the group of units of , letting refer to the additive group of invertible elements of . In addition, we write instead of for the set of atoms of the multiplicative monoid . Finally, for any such that divides in , we write instead of , and the term (greatest, maximal) common divisor in the context of semidomains are to be understood in the multiplicative monoid .
For the next example, we need the following lemma.
Lemma. For a semiring , the following conditions are equivalent.
(a) is a semidomain.
(b) The multiplication of extends to the Grothendieck group of turning into an integral domain.
Here is an educative example.
Example. Notice that the set is a monoid with the usual component-wise addition, and it is closed under the usual component-wise multiplication with multiplication identity . Hence is a semiring. Observe, on the other hand, that any extension of the multiplication of to making the latter a commutative ring must respect the identity and, therefore, will not turn into an integral domain. Hence it follows from the previous lemma that is not a semidomain.
(a) is a semidomain.
(b) The multiplication of extends to the Grothendieck group of turning into an integral domain.
(1) The polynomial extension is also a semidomain.
(2) The Laurent polynomial extension is a semidomain.
(3) The power series extension is a semidomain.
(4) If the additive monoid is reduced, then is a divisor-closed submonoid of .
Liao and Polo initiated the study of analogues of the Goldbach conjecture for polynomial semirings. Specifically, they showed that every that is not a monomial can be written as the sum of at most two irreducibles provided that . It is worth mentioning that the proof uses some facts about the distribution of prime numbers in . Note that this theorem is closely connected to the Goldbach conjecture. This result is about partitioning a fixed set of units into two subsets, where each subset represents an irreducible Laurent polynomial with positive integer coefficients, just as the Goldbach conjecture concerns the partitioning of a fixed set of units into two subsets, where each subset represents a positive prime. Moreover, if the Goldbach conjecture were true then, using that is irreducible when is a prime number, one could prove that Laurent polynomials for which is an even number greater than can be written as the sum of two irreducibles.
As noted in Resource , Hayes was the first to explore analogues of the Goldbach conjecture in the context of polynomial rings. In fact, he showed that polynomials with can be represented as the sum of two irreducibles. But, how does this statement relate to the Goldbach conjecture? For starters, in the context of Hayes' statement we are allowed to subtract. This somehow changes the nature of the Goldbach conjecture as elements might have various representations as sum of units. For instance, the polynomial can be also represented as . Observe that in the context of the original Goldbach conjecture is not a valid Goldbach decomposition of the number . On the other hand, assuming the Goldbach conjecture to be true does not provide much additional insight into Hayes's statement.
Polynomial Semidomains vs Laurent Polynomial Semidomains
As part of the exercises of Resource , you proved that and are semidomains. However, observe that no analogue of the Goldbach conjecture holds for . Indeed, since is an additive atom that is not a multiplicative unit, then the polynomial cannot be expressed as the sum of two irreducibles (see Exercise ). In contrast, in the context of we can write
where both expressions inside parentheses are irreducibles in . The key difference is that now the element is a multiplicative unit. For this reason, we focus on Laurent polynomial semidomains rather than polynomial semidomains.
One of the difficulties one encounters in establishing analogues of the Goldbach conjecture for polynomial semirings is finding suitable irreducibility criteria. Observe that, even though a semidomain can be nicely embedded into an integral domain , the irreducibles of does not coincide with the irreducibles of . For instance, the polynomial is irreducible in but, clearly, can be decomposed as when considered as a polynomial with coefficients in . Additionally, note that the units of and may differ. As a matter of fact, the series is a unit in but it is not a unit in . We can then conclude that the factorization properties of and are different in general.
Localization of Semidomains
Let be a semidomain, and let be a multiplicative subset of (i.e., a submonoid of ). Observe that is also a multiplicative subset of . Now set , where is the equivalence relation on defined as follows: if and only if (you will formally prove this in Exercise 2.1). We let denote the equivalence class of . Now define the following binary operations :
(i)
(ii) .
It is routine to verify that these operations are well defined and that is a semiring (we propose this as Exercise 2.2(a)). In fact, it is possible to prove that is a semidomain (this is Exercise 2.2(b)). The semidomain is called the localization of at the multiplicative set , and it is denoted by .
Let us take a look at a couple of simple examples of localization in the setting of semidomains. this construction seems very abstract!
Example 1: Consider the semidomain consisting of all nonnegative integers under the standard addition and. multiplication, and let denote the set of positive integers, which clearly a multiplicative subset of . What is ? Well, according to the previous construction, the elements of this new semidomain are fractions where and , and we say that provided that . This set with operations defined as in (i) and (ii) above looks a lot like the semidomains consisting of all nonnegative rational numbers. According to Exercise 2.3, this semidomain is .
Example 2: Let us start once again with the semidomain , but now fix a prime and consider the subset . Clearly, is a multiplicative subset of the semidomain , and it is not hard to see that the elements of are fractions with being a nonnegative integer and being a nonnegative power of . As it turns out, this semidomain is isomorphic to , we it is stated in Exercise 2.4.
Observe that the elements of are fractions of the form , with and . Moreover, it is not hard to show that all elements of the form with are units in . As a matter of fact, this is the ``simplest extension" of the semidomain for which all elements of become units (this is the order of Exercise 2.5).
(b) Show that is actually a semidomain.
(i) ;
(ii) .
If, additionally, the map is a bijection (i.e., simultaneously injective and surjective), then we say that is a semiring isomorphism and also that the semirings and are isomporphic.
(a) Prove that being isomorphic determines an equivalence relation in the class consisting of all semirings.
(b) Prove that the semidomain defined in Example 1 is isomorphic to the semiring .
(a) The map given by is an injective semiring homomorphism satisfying that is a unit in for all .
(b) If is a semiring homomorphism such that is a unit in for every , then there exists a unique semiring homomorphism such that .